# Let G be a simple connected planar graph with at least three vertices. Prove that if every face of G has degree 3, then G has an even number of faces

Let G be a simple connected planar graph with at least three vertices. Prove that if every face of G has degree 3, then G has an even number of faces

I've looked around and haven't found any questions similar to this.

I'm trying to prove this. I know that every face shares two edges so if I were to use Euler's formula 3f = 2m. Im not really sure where to go from there.

• You should use Euler formula, and the fact that $3f$ gives you twice the number of edges because each edge is counted twice. From there, you simply need to argue that the number of vertices is an integer. Commented May 22, 2018 at 8:27
• So would it be something like since 3f = 2m, m = 3/2f so f has to be even? Commented May 22, 2018 at 8:32
• Exactly. Otherwise, plugging that value for $m$ in Euler's formula, you find that $n-2$ is not an integer. Commented May 22, 2018 at 8:34
• I get what your saying, does this mean that n - 3/2f + f = 2. which only true for even faces Commented May 22, 2018 at 8:36
• Yes, that's the proof. Commented May 22, 2018 at 12:41

I know that every face shares two edges so if I were to use Euler's formula $3f = 2m$
What's $m$? What does Euler's formula have to do with it? In fact, being planar is largely irrelevant: the same is true on surfaces of arbitrary genus.
If every face has degree $3$, every face has $3$ edges; but every edge has two faces, so $E = \frac{3}{2}F$ where $E$ is the number of edges and $F$ is the number of faces; then $2E = 3F$ and by the fundamental theorem of arithmetic $2$ must be a factor either of $3$ (which it isn't) or of $F$.